Confidence Interval Worksheet (6



Confidence Interval Worksheet (6.1)

1) Apartment rental rates. You want to rent an unfurnished one-bedroom apartment for next semester. The mean monthly rent for a random sample of 10 apartments advertised in the local newspaper is $540. Assume that the standard deviation is $80. Find a 95% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community.

Compare the margin of error for intervals with 90, 95, and 99% confidence:

2) Clothing for runners. Your company sells exercise clothing and equipment on the Internet. To design the clothing, you collect data on the physical characteristics of your different types of customers. Here are the weights (in kilograms) for a sample of 24 male runners. Suppose the standard deviation of the population is known to be 4.5 kg.

67.8 61.9 63.0 53.1 62.3 59.7 55.4 58.9 60.9 69.2 63.7 68.3 64.7 65.6 56.0 57.8 66.0 62.9 53.6 65.0 55.8 60.4 69.3 61.7

(a) What is the standard deviation of x-bar?

(b) Give a 95% confidence interval for µ, the mean of the population from which the sample is drawn.

(The sample mean is 61.79, to save you some time).

(c) Will the interval contain the weights of approximately 95% of all similar runners? Explain.

Suppose a 25th male runner had a reported weight of 92.3 kg. Find the 95% interval for the new data and compare.

(Just think about this. I DO expect that you should know HOW to do this, although you don’t have to do the calculations here, because they won’t fit in this itty-bitty space).

3) 99% versus 95% confidence interval. Find a 99% confidence interval for the mean weight µ of the population of male runners in #2 (with n=24). Is the 99% confidence interval wider or narrower than the 95% interval found in Exercise 6.12? Explain in plain language why this is true.

Suppose you want the 99% interval to be the same width as the 95% interval. How many additional runners would you need to include in your sample?

Confidence Interval Worksheet (6.1)

1) Apartment rental rates. You want to rent an unfurnished one-bedroom apartment for next semester. The mean monthly rent for a random sample of 10 apartments advertised in the local newspaper is $540. Assume that the standard deviation is $80. Find a 95% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community.

Compare the margin of error for intervals with 90, 95, and 99% confidence:

2) Clothing for runners. Your company sells exercise clothing and equipment on the Internet. To design the clothing, you collect data on the physical characteristics of your different types of customers. Here are the weights (in kilograms) for a sample of 24 male runners. Suppose the standard deviation of the population is known to be 4.5 kg.

67.8 61.9 63.0 53.1 62.3 59.7 55.4 58.9 60.9 69.2 63.7 68.3 64.7 65.6 56.0 57.8 66.0 62.9 53.6 65.0 55.8 60.4 69.3 61.7

(a) What is the standard deviation of x-bar?

(b) Give a 95% confidence interval for µ, the mean of the population from which the sample is drawn.

(The sample mean is 61.79, to save you some time).

(c) Will the interval contain the weights of approximately 95% of all similar runners? Explain.

Suppose a 25th male runner had a reported weight of 92.3 kg. Find the 95% interval for the new data and compare.

(Just think about this. I DO expect that you should know HOW to do this, although you don’t have to do the calculations here, because they won’t fit in this itty-bitty space).

3) 99% versus 95% confidence interval. Find a 99% confidence interval for the mean weight µ of the population of male runners in #2 (with n=24). Is the 99% confidence interval wider or narrower than the 95% interval found in Exercise 6.12? Explain in plain language why this is true.

Suppose you want the 99% interval to be the same width as the 95% interval. How many additional runners would you need to include in your sample?

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